Is it possible, that if a 1 < a 2 < a 3 < a 4 < ⋯ < a 2 0 1 7 are positive integers, then a 1 1 − a 2 1 = a 2 1 − a 3 1 = a 3 1 − a 4 1 = ⋯ = a 2 0 1 6 1 − a 2 0 1 7 1 ?
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Define the sequence a n by the following:
a n = N − n k , 1 ≤ n < N
where k and N are fixed positive integers.
We'll prove that for some appropiate k and N this sequence fulfills the problem's conditions.
First, we seek to show that the difference a i 1 − a i + 1 1 does not depend on the value of i . Indeed,
a i 1 − a i + 1 1 = k N − i − k N − i − 1 = k 1 .
Letting k = N ! it follows that N − n ∣ k for every 1 ≤ n < N , hence the value of a n = N − n k is a positive integer. Since the condition a i < a i + 1 is the simple consequence of the fact that N − i k < N − i − 1 k , we have shown that this sequence is a correct solution for any N . Putting in N = 2 0 1 8 we are done.